Question: Suppose we wanted to evaluate the double integral $S = \iint_D 2x - 2y \, dx \, dy$ by first applying a change of variables from $D$ to $R$ : $\begin{aligned} x &= X_1(u, v) = \dfrac{u}{2} - \dfrac{v}{3} \\ \\ y &= X_2(u, v) = \dfrac{u}{2} + \dfrac{v}{3} \end{aligned}$ What is $S$ under the change of variables? If you know an expression within absolute value is non-negative, do not use absolute value at all. $S = \iint_R $ $ du \, dv$
Solution: If we have a transformation $\bold{X} : R \to D$, then we can rewrite an integral under the change of variables: $ \iint_D f(x, y) \, dA = \iint_R f(\bold{X}(u, v)) | J(\bold{X}) | \, du \, dv$ First we need to find the absolute value of the Jacobian, $|J(\bold{X})|$. $\begin{aligned} |J(\bold{X})| &= \left| \det \begin{pmatrix} \dfrac{\partial X_1}{\partial u} & \dfrac{\partial X_1}{\partial v} \\ \\ \dfrac{\partial X_2}{\partial u} & \dfrac{\partial X_2}{\partial v} \end{pmatrix} \right| \\ \\ &= \left| \det \begin{pmatrix} \dfrac{1}{2} & \dfrac{-1}{3} \\ \\ \dfrac{1}{2} & \dfrac{1}{3} \end{pmatrix} \right| \\ \\ &= \left| \dfrac{1}{6} + \dfrac{1}{6} \right| \\ \\ &= \dfrac{1}{3} \end{aligned}$ Now we substitute $u$ and $v$ in $f(x, y)$. $\begin{aligned} f(x, y) &= f(\bold{X}(u, v)) \\ \\ &= 2 X_1(u, v) - 2 X_2(u, v) \\ \\ &= u - \dfrac{2v}{3} - \left( u + \dfrac{2v}{3} \right) \\ \\ &= \dfrac{-4v}{3} \end{aligned}$ Putting everything together, we get the integral under the change of variables: $ \iint_R \dfrac{-4v}{9} \, du \, dv$